A) \[1\mu F\]
B) \[2\mu F\]
C) \[\frac{1}{2}\mu F\]
D) \[\infty \]
Correct Answer: B
Solution :
This combination forms a G.P. \[S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\] Sum of infinite G.P., \[S=\frac{a}{1-r}\] Here a = first term = 1 and r = common ratio\[=\frac{1}{2}\] \[\therefore \] \[S=\frac{1}{1-\frac{1}{2}}=2\] \[\Rightarrow \] \[{{C}_{eq}}=2\mu F\]
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